If two charges +q and +4q are separated by a distance ‘d’ and a point charge Q is placed on the line joining the above two charges and in between them such that all charges are in equilibrium. Then the charge Q and it’s position are

Charges = q and 4q distance = d
third charge = Q
distance = x from charge q

$$\begin{gathered} \frac{\mathrm{kq}}{{\mathrm{x}}^2}=\frac{\mathrm{k}\left(4\mathrm{q}\right)}{(\mathrm{d}-\mathrm{x}{)}^2} \\ \Rightarrow \frac{(\mathrm{d}-\mathrm{x}{)}^2}{{\mathrm{x}}^2} = 4 \\ \Rightarrow \frac{(\mathrm{d} - \mathrm{x})}{\mathrm{x}} = 2 \mathrm{or} \frac{{\displaystyle (\mathrm{d} - \mathrm{x})}}{{\displaystyle \mathrm{x}}} = -2 \\ \Rightarrow \mathrm{x} = \frac{\mathrm{d}}{3} \mathrm{or} \mathrm{x} = -\mathrm{d} \end{gathered}$$

The force between +q charge and third charge should be of attractive nature. Thus the nature of the 3rd charge is (−ve).
Taking the third charge to be −Q (say) and then on applying the condition of equilibrium on +q charge:

$\frac{\mathrm{KQ}}{{\mathrm{x}}^2}+\frac{\mathrm{K}\left(4\mathrm{q}\right)}{{\mathrm{d}}^2} = 0$

$\Rightarrow \frac{\mathrm{KQ}}{(\mathrm{d}/3{)}^2}+\frac{\mathrm{K}\left(4\mathrm{q}\right)}{{\mathrm{d}}^2} = 0$

$\mathrm{putting} \mathrm{equations}:$

$\begin{array}{r}\Rightarrow \frac{9\mathrm{KQ}}{{\mathrm{d}}^2}+\frac{4\mathrm{Kq}}{{\mathrm{d}}^2} = 0\\ \Rightarrow 9\mathrm{Q}=-4\mathrm{q} \\ \text{⇒ }\mathrm{Q}=-\frac{4q}{9} \end{array}$